Certainly! Here’s a structured two-step answer with explanations: --- ### Step 1: Conservation of Energy on Frictionless Inclined Planes **Explanation:** For frictionless inclined planes with the same vertical height, the only force doing work on the object is gravity. When the object is released from the top, it loses potential energy and gains kinetic energy. The total mechanical energy is conserved. The vertical height determines the loss in potential energy, not the length or angle of the incline. --- ### Step 2: Final Speed at the Bottom **Explanation:** Since the vertical height (h) is the same for both planes, the change in potential energy (\(\Delta PE = mgh\)) is the same. By energy conservation: \[ mgh = \frac{1}{2}mv^2 \] Solving for \(v\) gives: \[ v = \sqrt{2gh} \] So, the speed at the bottom depends only on the vertical height, not on the path taken. --- ## Final Answer: **The object will have the same speed at the bottom of both frictionless inclined planes with the same vertical height, regardless of the path or angle of the incline.**